The Integral Test — AP® Calculus BC Lesson

Part 1 · Hook

Why this matters

Stack boxes of width 1 and heights

\frac{1}{4}, \frac{1}{9}, \frac{1}{16}, \ldots

along the

x

-axis, and they fit neatly under the curve

y = \frac{1}{x^2}

. If the area under the curve is finite, the boxes, which are the terms of a series, cannot add up to infinity. That picture is the integral test: it trades a series you cannot add for an improper integral you can evaluate.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. $\int_1^{\infty} \frac{1}{x^2}\,dx$ equals

$1$
$\infty$
$\frac{1}{2}$
$0$

Show the answer

$\lim_{b \to \infty} \left(1 - \frac{1}{b}\right)$ $= 1$ .

Correct: $1$ :
$\infty$ :
$\frac{1}{2}$ :
$0$ :

2. $f$ is decreasing on an interval when

$f'(x) < 0$ there
$f(x) < 0$ there
$f''(x) < 0$ there
$f'(x) = 0$ there

Show the answer

A negative derivative means the function falls as $x$ grows.

Correct: $f'(x) < 0$ there:
$f(x) < 0$ there:
$f''(x) < 0$ there:
$f'(x) = 0$ there:

Part 4 · See it

See it first

Left: the curve f(x) = 1/x² with boxes of width 1 drawn under it; box n spans x = n − 1 to n and has height 1/n², for n = 2 to 7. The boxes fit under the curve from x = 1 on, so 1/2² + 1/3² + ... is less than the integral from 1 to infinity of 1/x², which is 1: the series converges. Right: the curve f(x) = 1/x with boxes of height 1/n from x = n to n + 1, for n = 1 to 6, sticking out above the curve, so 1 + 1/2 + ... + 1/N is more than the integral from 1 to N + 1 of 1/x, which is ln(N + 1) and grows without bound: the series diverges. — Boxes of height $\frac{1}{n^2}$ fit under $y = \frac{1}{x^2}$ , whose area is finite; boxes of height $\frac{1}{n}$ stick out above $y = \frac{1}{x}$ , whose area is infinite. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Match the terms with a function: $a_n = f(n)$ , with $f$ positive, continuous and decreasingeach term is the area of a box of width 1 that fits beside the curve
Boxes under the curve, from $x = 1$ on $a_2$ $+ a_3$ $+ \cdots \le \int_1^{\infty} f(x)\,dx$ : a finite integral holds the series down
Boxes over the curve $a_1$ $+ \cdots$ $+ a_n \ge \int_1^{n+1} f(x)\,dx$ : an infinite integral pushes the series up
So the series and the integral behave alikeboth converge or both diverge

Part 6 · Key ideas

Key ideas

Integral test: if $f$ is positive, continuous and decreasing on $[k, \infty)$ and $a_n = f(n)$ , then $\sum_{n=k}^{\infty} a_n$ and $\int_k^{\infty} f(x)\,dx$ both converge or both diverge.
State the three conditions every time. \"Decreasing\" usually needs $f'(x) < 0$ ; it may hold only from some $x$ on, and then the test starts there.
The integral's value is not the series' sum. The test decides convergence only.
Write the improper integral as a limit, $\lim_{b \to \infty} \int_k^{b} f(x)\,dx$ , to the end.

Part 7 · Misconception

A common mistake

The wrong idea: $\int_1^{\infty} \frac{dx}{x^2}$ $= 1$ , so $\sum_{n=1}^{\infty} \frac{1}{n^2}$ $= 1$ .

What actually happens: The integral test shows the series converges, nothing more. Its first term is already 1, and the sum is about 1.645.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Four series and their improper integrals

Each series, the function $f$ with $f(n) = a_n$ , and the matching improper integral (each $f$ is positive, continuous and decreasing on the interval used)
Series	$f(x)$	Improper integral
$\sum_{n=1}^{\infty} \frac{1}{n^2}$	$\frac{1}{x^2}$	$\int_1^{\infty} \frac{dx}{x^2}$ $= 1$
$\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}$	$\frac{1}{x^2 + 1}$	$\int_1^{\infty} \frac{dx}{x^2 + 1}$ $= \frac{\pi}{4}$
$\sum_{n=2}^{\infty} \frac{1}{n\ln n}$	$\frac{1}{x\ln x}$	$\int_2^{\infty} \frac{dx}{x\ln x}$ diverges
$\sum_{n=1}^{\infty} \frac{n}{n^2 + 1}$	$\frac{x}{x^2 + 1}$	$\int_1^{\infty} \frac{x\,dx}{x^2 + 1}$ diverges

1. What does the integral test say about $\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}$ ?

It converges, since its integral is finite
It converges to $\frac{\pi}{4}$
It diverges, since $\frac{\pi}{4}$ is not 0
Nothing, since the integral is not a whole number

Show the answer

$f(x) = \frac{1}{x^2 + 1}$ is positive, continuous and decreasing for $x \ge 1$ , and $\int_1^{\infty} f(x)\,dx$ $= \frac{\pi}{4}$ is finite, so the series converges by the integral test.

Correct: It converges, since its integral is finite: Right: a finite improper integral means the series converges.
It converges to $\frac{\pi}{4}$ : The test gives convergence, not the sum. The sum is not $\frac{\pi}{4}$ .
It diverges, since $\frac{\pi}{4}$ is not 0: That mixes in the nth term test, which is about the limit of the terms, not the integral.
Nothing, since the integral is not a whole number: Any finite value of the integral gives convergence.

2. A student concludes from the table that $\sum_{n=1}^{\infty} \frac{1}{n^2}$ $= 1$ . Is that right?

Yes: the integral test gives the sum
No: the series diverges, though the integral is finite
Yes: both are limits of sums of $\frac{1}{x^2}$
No: the integral decides convergence but is not the sum

Show the answer

The integral test says only whether the series converges. Here $a_1 = 1$ and the other terms are positive, so the sum is more than 1 (it is $\frac{\pi^2}{6} \approx 1.645$ ).

Yes: the integral test gives the sum: The test compares the series with the integral; the two are not equal.
No: the series diverges, though the integral is finite: The integral is finite, so the series converges; it just does not equal 1.
Yes: both are limits of sums of $\frac{1}{x^2}$ : The integral adds areas under a curve; the series adds boxes of height $\frac{1}{n^2}$ , which differ.
Correct: No: the integral decides convergence but is not the sum: Right: the sum is more than 1, since its first term alone is 1.

3. What does the integral test say about $\sum_{n=2}^{\infty} \frac{1}{n\ln n}$ ?

It converges, since its terms approach 0
Nothing, since the series starts at $n = 2$
It diverges, since its integral diverges
It converges, since $n\ln n$ grows faster than $n$

Show the answer

With $u = \ln x$ , $\int \frac{dx}{x\ln x}$ $= \ln|\ln x| + C$ . So $\int_2^{b} \frac{dx}{x\ln x}$ $= \ln(\ln b) - \ln(\ln 2) \to \infty$ : the integral diverges, so the series diverges by the integral test.

It converges, since its terms approach 0: Terms approaching 0 never prove convergence; the integral decides here.
Nothing, since the series starts at $n = 2$ : The test works from any starting index; integrate from that index.
Correct: It diverges, since its integral diverges: Right: $\int_2^b \frac{dx}{x\ln x}$ $= \ln(\ln b) - \ln(\ln 2)$ grows without bound.
It converges, since $n\ln n$ grows faster than $n$ : Growing faster than $n$ is not enough; the integral grows like $\ln(\ln b)$ , without bound.

Table

The function $f(x) = xe^{-x}$

Facts about $f(x) = xe^{-x}$ for the series $\sum_{n=1}^{\infty} ne^{-n}$
Quantity	Formula
$f(x)$	$xe^{-x}$
$f'(x)$	$(1 - x)e^{-x}$
Series	$\sum_{n=1}^{\infty} ne^{-n}$

4. On which interval is $f$ decreasing?

$x \ge 0$
$x \le 1$
$x \ge e$
$x \ge 1$

Show the answer

$e^{-x} > 0$ , so the sign of $f'(x)$ is the sign of $1 - x$ , which is negative for $x > 1$ . $f$ is decreasing on $[1, \infty)$ , which is where the series starts.

$x \ge 0$ : For $0 < x < 1$ , $1 - x > 0$ , so $f' > 0$ and $f$ is increasing there.
$x \le 1$ : That is where $f' \ge 0$ , so $f$ is increasing there.
$x \ge e$ : $f'$ changes sign at $x = 1$ , where $1 - x = 0$ ; $e^{-x}$ is never 0.
Correct: $x \ge 1$ : Right: $f'(x)$ $= (1 - x)e^{-x} \le 0$ when $x \ge 1$ .

5. Find $\int_1^{\infty} xe^{-x}\,dx$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$\lim_{b \to \infty} \Big[-xe^{-x} - e^{-x}\Big]_1^{b}$ $= \lim_{b \to \infty} \left(-\frac{b + 1}{e^{b}}\right)$ $+ \frac{2}{e}$ $= 0$ $+ \frac{2}{e}$ $= \frac{2}{e}$ (L'Hospital's rule gives $\frac{b + 1}{e^b} \to 0$ ).

Answer: $\frac{2}{e}$ ≈ 0.736

6. Which conditions must $f$ meet on $[k, \infty)$ for the integral test to apply to $\sum_{n=k}^{\infty} a_n$ with $a_n = f(n)$ ?

Positive, continuous and increasing
Positive, continuous and decreasing
Continuous, with $\lim_{x \to \infty} f(x) = 0$
Differentiable and bounded

Show the answer

The integral test needs $f$ positive, continuous and decreasing on $[k, \infty)$ ; then $\sum a_n$ and $\int_k^{\infty} f(x)\,dx$ both converge or both diverge.

Positive, continuous and increasing: An increasing positive $f$ gives terms that do not approach 0; the test needs decreasing.
Correct: Positive, continuous and decreasing: Right: all three, on the interval where the series runs.
Continuous, with $\lim_{x \to \infty} f(x) = 0$ : A limit of 0 is not one of the conditions, and it is not enough.
Differentiable and bounded: These are not the test's conditions; positive, continuous and decreasing are.

Part 9 · Summary

Summary

For a series of positive terms

a_n = f(n)

with

f

positive, continuous and decreasing, compute

\lim_{b \to \infty} \int_k^{b} f(x)\,dx

. A finite limit means the series converges; an infinite one means it diverges. Check the conditions, keep limit notation, and never report the integral as the sum.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

Four series and their improper integrals

The function f(x)=xe−xf(x) = xe^{-x}

Summary

What comes next

Connections

The function $f(x) = xe^{-x}$